A particle of charge 1.6 μC and mass 16 μg is fired perpendicular to a uniform magnetic field of magnitude 6.28 T. Find the time in seconds taken by the particle to return to its original position for the first time.
Physics · Electrostatics · Electric field intensity
JEE Mains 2025 4 April Evening Shift

Question

A particle of charge 1.6 μC and mass 16 μg is fired perpendicular to a uniform magnetic field of magnitude 6.28 T. Find the time in seconds taken by the particle to return to its original position for the first time.

Physics · Electrostatics · Electric field intensity

JEE Mains 2025 4 April Evening Shift

G1 Physics · Accepted Answer

A charged particle moving perpendicular to a uniform magnetic field executes uniform circular motion. Its cyclotron period is T = 2πm/(|q|B). Convert units: m = 16 μg = 16×10⁻⁹ kg, q = 1.6 μC = 1.6×10⁻⁶ C, B = 6.28 T, π = 3.14. Therefore, T = [2(3.14)(16×10⁻⁹)]/[(1.6×10⁻⁶)(6.28)]. Since 2π = 6.28, T = (6.28×16×10⁻⁹)/(1.6×10⁻⁶×6.28) = (16/1.6)×10⁻³ = 10×10⁻³ = 0.01 s. The particle first returns to the starting point after one complete revolution. Answer: 0.01.